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Another integral problem

paulmdrdo

Active member
May 13, 2013
386
∫(tan3x/sec2x)dx

this is what i do:

∫(tan3xsec-2x)dx
∫(tan2xtanxsec-1xsec-1x)dx
∫(sec2x-1)tanxsec-1xsec-1x)dx
let u = secx;
du = tanxsecxdx

now i'm stuck! please help.:confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: another integ problem.

I think my approach would be to rewrite the integrand as follows:

\(\displaystyle \frac{\tan^3(x)}{\sec^2(x)}=\frac{\sin^3(x)}{\cos^3(x)}\cos^2(x)=\frac{-\sin(x)\left(\cos^2(x)-1 \right)}{\cos(x)}\)

Now you should see a $u$-substitution that will work quite nicely.
 

paulmdrdo

Active member
May 13, 2013
386
Re: another integ problem.

I think my approach would be to rewrite the integrand as follows:

\(\displaystyle \frac{\tan^3(x)}{\sec^2(x)}=\frac{\sin^3(x)}{\cos^3(x)}\cos^2(x)=\frac{-\sin(x)\left(\cos^2(x)-1 \right)}{\cos(x)}\)

Now you should see a $u$-substitution that will work quite nicely.
i used that technique makfl and i get the right answer. but for educational purposes i want to know another way of solving that.

this what i do a while ago.

∫(tan3x/sec2x)dx
∫(tan2xtanx/sec2x)dx

∫(sec2x-1)tanx)/sec2x)dx
∫(sec2xtanx-tanx)/sec2x)dx
∫[(sec2xtanx)/sec2x - tanx/sec2x]dx
∫tanxdx - ∫(tanx/sec2x)dx

ln|secx| - ∫(sinx/cosx*cos2x)dx
ln|secx| - ∫sinxcosxdx

u= sinx; du = cosxdx

ln|secx| - ∫udu
ln|secx| - u2/2
ln|secx| - sin2x/2 + C

is my solution correct..it is different with the one i get when i used the technique you suggested.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: another integral problem.

Yes, your solution is also correct, and many times when we find differing forms for an indefinite integral, we find that they (without the constant of integration) differ only by a constant, which of course is allowed.

So, let's look at the two solutions without the constants:

(1) \(\displaystyle \frac{1}{2}\cos^2(x)-\ln|\cos(x)|\)

(2) \(\displaystyle \ln|\sec(x)|-\frac{1}{2}\sin^2(x)\)

First, we should observe that:

\(\displaystyle \ln|\sec(x)|=\ln|(\cos(x))^{-1}|=-\ln|\cos(x)|\)

So, if we subtract one solution from the other, we should obtain a constant. Subtracting (2) from (1), we get:

\(\displaystyle \frac{1}{2}\cos^2(x)+\frac{1}{2}\sin^2(x)\)

Using a Pythagorean identity, we now have:

\(\displaystyle \frac{1}{2}\)

So, we see that the two solutions differ by a constant, which means they are equivalent as an anti-derivative.

Perhaps a simpler way to look at it is to take the solution you obtained, and rewrite it:

\(\displaystyle \ln|\sec(x)|-\frac{1}{2}\sin^2(x)+C=-\ln|\cos(x)|-\frac{1}{2}\left(1-\cos^2(x) \right)+C=\frac{1}{2}\cos^2(x)-\ln|\cos(x)|+\left(C-\frac{1}{2} \right)\)

Since an arbitrary constant less one half, is still an arbitrary constant, we get:

\(\displaystyle \frac{1}{2}\cos^2(x)-\ln|\cos(x)|+C\)
 

paulmdrdo

Active member
May 13, 2013
386
Re: another integral problem.

Yes, your solution is also correct, and many times when we find differing forms for an indefinite integral, we find that they (without the constant of integration) differ only by a constant, which of course is allowed.

So, let's look at the two solutions without the constants:

(1) \(\displaystyle \frac{1}{2}\cos^2(x)-\ln|\cos(x)|\)

(2) \(\displaystyle \ln|\sec(x)|-\frac{1}{2}\sin^2(x)\)

First, we should observe that:

\(\displaystyle \ln|\sec(x)|=\ln|(\cos(x))^{-1}|=-\ln|\cos(x)|\)

So, if we subtract one solution from the other, we should obtain a constant. Subtracting (2) from (1), we get:

\(\displaystyle \frac{1}{2}\cos^2(x)+\frac{1}{2}\sin^2(x)\)

Using a Pythagorean identity, we now have:

\(\displaystyle \frac{1}{2}\)

So, we see that the two solutions differ by a constant, which means they are equivalent as an anti-derivative.

Perhaps a simpler way to look at it is to take the solution you obtained, and rewrite it:

\(\displaystyle \ln|\sec(x)|-\frac{1}{2}\sin^2(x)+C=-\ln|\cos(x)|-\frac{1}{2}\left(1-\cos^2(x) \right)+C=\frac{1}{2}\cos^2(x)-\ln|\cos(x)|+\left(C-\frac{1}{2} \right)\)

Since an arbitrary constant less one half, is still an arbitrary constant, we get:

\(\displaystyle \frac{1}{2}\cos^2(x)-\ln|\cos(x)|+C\)
i just want to ask how is ln|secx| = -ln|cosx|... the rest of explanation is enligthening. thanks a lot!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: another integral problem.

Recall the logarithmic property:

\(\displaystyle \log_a\left(b^c \right)=c\cdot\log_a(b)\)

and the definition:

\(\displaystyle \sec(x)\equiv\frac{1}{\cos(x)}=(\cos(x))^{-1}\)
 

soroban

Well-known member
Feb 2, 2012
409
Re: another integ problem.

Hello, paulmdrdo!

The integral [tex]\int \sin x\cos x\,dx[/tex] is famous
for having different (but equivalent) answers.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[tex]1.\: \int \sin x(\cos x\,dx)[/tex]

Let [tex]u \,=\,\sin x \quad\Rightarrow\quad du \,=\,\cos x\,dx[/tex]

Substitute: .[tex]\int u\,du \:=\:\tfrac{1}{2}u^2+C[/tex]

Back-substitute: .[tex]\tfrac{1}{2}\sin^2x + C[/tex]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[tex]2.\;\int \cos x(\sin x\,dx)[/tex]

Let [tex]u = \cos x \quad\Rightarrow\quad du = \text{-}\sin x\,dx \quad\Rightarrow\quad \sin x\,dx = \text{-}du[/tex]

Substitute: .[tex]\int u(\text{-}du) \:=\:\text{-}\int u\,du \:=\:\text{-}\tfrac{1}{2}u^2+C[/tex]

Back-substitute: .[tex]\text{-}\tfrac{1}{2}\cos^2x + C[/tex]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[tex]3.\;\int\sin x\cos x\,dx \:=\:\tfrac{1}{2}\int2\sin x\cos x\,dx [/tex]

. . . [tex]=\;\tfrac{1}{2}\int\sin2x\,dx \:=\:\text{-}\tfrac{1}{4}\cos2x + C[/tex]